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Solve the differential equation. $$\frac { d y } { d \theta } = \frac { e ^ { y } \sin ^ { 2 } \theta } { y \sec \theta }$$

Short Answer

Expert verified
The differential equation solves to \(-ye^{-y} + e^{-y} = \frac{\sin^3 \theta}{3} + C\).

Step by step solution

01

Simplify the Equation

Start by simplifying the given differential equation: \( \frac{dy}{d\theta} = \frac{e^y \sin^2 \theta}{y \sec \theta} \). Notice that \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, the equation can be rewritten as \( \frac{e^y \sin^2 \theta \cos \theta}{y} \).
02

Separate Variables

Separate the variables by bringing all terms involving \( y \) to one side and all terms involving \( \theta \) to the other side. This gives us \( y \frac{dy}{e^y} = \sin^2 \theta \cos \theta \, d\theta \).
03

Integrate Both Sides

Integrate both sides of the equation to solve for \( y \) in terms of \( \theta \). The left side integrates to \( \int y e^{-y} \, dy \) and the right side integrates to \( \int \sin^2 \theta \cos \theta \, d\theta \).
04

Integrate Left Side

Use integration by parts for the left side: let \( u = y \) and \( dv = e^{-y} \, dy \). Then, \( du = dy \) and \( v = -e^{-y} \). The integral becomes \( -ye^{-y} - \int -e^{-y} \, dy = -ye^{-y} + e^{-y} \).
05

Integrate Right Side

Use substitution for the right side: let \( u = \sin \theta \), so \( du = \cos \theta \, d\theta \), and the integral becomes \( \int u^2 \, du = \frac{u^3}{3} = \frac{(\sin \theta)^3}{3} \).
06

Equate and Solve

Equate the integrated results: \( -ye^{-y} + e^{-y} = \frac{\sin^3 \theta}{3} + C \), where \( C \) is the constant of integration. This is the general solution.
07

Optional - Solve for Specific Solutions

You may solve explicitly for \( y \) if initial conditions are provided, but without particular conditions, the solution remains in implicit form as above.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
In the world of calculus, mastering integration techniques is key to solving complex differential equations. Integration is the process of finding the integral or antiderivative of a function, which is essentially the reverse process of differentiation. Different integration techniques are like tools in a toolbox, helping us solve various types of problems.

Some common techniques include:
  • Substitution: Used when an integral contains a composition of functions, making it easier to evaluate by changing variables.
  • Integration by parts: This is used when the product of two functions is present, allowing us to break down integrals into simpler parts.
  • Partial fraction decomposition: Useful for rational functions, breaking them into simpler fractions to integrate more easily.
In our problem, multiple integration techniques are used to solve different parts of the equation effectively. This stepwise approach shows the flexibility of integration methods in problem-solving, allowing us to tackle complex expressions in manageable pieces.
Variable Separation
One of the key methods for solving differential equations is variable separation, a technique where we separate the variables of the equation to simplify integration. This means all terms involving one variable, such as \( y \), are moved to one side of the equation. Meanwhile, all terms involving the other variable, like \( \theta \) in our case, are moved to the opposite side.

The goal is to have an equation in the form:
  • \( f(y) \, dy = g(\theta) \, d\theta \)
This makes it easier to integrate both sides separately. In our exercise, we simplified the original differential equation to \( y \frac{dy}{e^y} = \sin^2 \theta \, \cos \theta \, d\theta \). This clear separation allows us to focus on integrating one variable at a time.
By separating the variables, we simplify the problem and make it more approachable, especially when dealing with intricate expressions in differential equations.
Trigonometric Substitution
Trigonometric substitution is an integration technique useful when dealing with non-polynomial functions involving trigonometric expressions. This method leverages trigonometric identities to simplify integrals and make them solvable.

In our exercise, the right side of the equation \( \int \sin^2 \theta \, \cos \theta \, d\theta \) was simplified using substitution:
  • We set \( u = \sin \theta \) which simplifies the integral to \( \int u^2 \, du \).
  • This reduces the complexity of the expression, transforming it into more straightforward polynomial integration, \( \frac{u^3}{3} \).
This substitution helps tackle intricate trigonometric integrals by converting them into simple polynomial forms. It serves as a powerful tool to simplify integrals that otherwise might seem daunting.
Integration by Parts
Integration by parts comes in handy when you have a product of two functions you're trying to integrate. This technique breaks down the integral into simpler parts by using the formula:
  • \( \int u \, dv = uv - \int v \, du \)
By strategically choosing \( u \) and \( dv \), you transform complex integrals into easier ones.

In our case, for the left side of the differential equation, we employ integration by parts:
  • Let \( u = y \) and \( dv = e^{-y} \, dy \), leading to \( du = dy \) and \( v = -e^{-y} \).
This transforms the integral into \( -ye^{-y} + \int e^{-y} \, dy = -ye^{-y} + e^{-y} \).
This method shows how integration by parts can simplify products of algebraic and exponential functions, making it vital for solving differential equations efficiently.

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Most popular questions from this chapter

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. \(y^{\prime}=x+y^{2}, \quad(0,0)\)

13-15 Drug dissolution Differential equations have been used extensively in the study of drug dissolution for patients given oral medications. The three simplest equations used are the zero-order kinetic equation, the Noyes-Whitney equation, and the Weibull equation. All assume that the initial concentration is zero but make different assumptions about how the concentration increases over time during the dissolution of the medication. The zero-order kinetic equation states that the rate of change in the concentration of drug \(c(\) in \(\mathrm{mg} / \mathrm{mL})\) during dissolution is governed by the differential equation $$\frac{d c}{d t}=k$$ where \(k\) is a positive constant. Is this differential equation pure-time, autonomous, or nonautonomous? State in words what this differential equation says about how drug dissolution occurs. What is the solution of this differential equation with the initial condition \(c(0)=0 ?\)

\(11-18\) Find the solution of the differential equation that satisfies the given initial condition. $$\frac { d y } { d x } = \frac { x } { y } , \quad y ( 0 ) = - 3$$

(a) Program a calculator or computer to use Euler's method to compute \(y(1),\) where \(y(x)\) is the solution of the initial-value problem $$\frac{d y}{d x}+3 x^{2} y=6 x^{2} \quad y(0)=3$$ for each of the given step sizes. \(\begin{aligned} \text { (i) } h=1 & \text { (ii) } h=0.1 \\ \text { (iii) } h=0.01 & \text { (iv) } h=0.001 \end{aligned}\) (b) Verify that \(y=2+e^{-x^{3}}\) is the exact solution of the differential equation. (c) Find the errors in using Euler's method to compute \(y(1)\) with the step sizes in part (a). What happens to theerror when the step size is divided by 10\(?\)

.Find the equilibria for the differential equation and determine the values of \(a\) for which each equilibrium is locally stable, Assume \(a \neq 0\) . (a) $$y^{\prime}=1+a y$$ (b) $$y^{\prime}=1-e^{-a y}$$ (c) \(y^{\prime}=a e^{y} \cos y, \quad 0

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